Question

4.52 Random variables X and Y follow a joint distribution f(x, y) = 2, 0 < x ≤ y < 1, 0, otherwise. Determine the correlation coefficient between X and Y .

Answer 1

First, you'll need to find the marginal distributions of $X,Y$. By the law of total probability,

$P(X=x)=\displaystyle\sum_yP(X=x)P(Y=y)$

which translates to

$f_X(x)=\displaystyle\int_x^1f_{X,Y}(x,y)\,\mathrm dy=\begin{cases}2(1-x)&\text{for }0$

Similarly,

$f_Y(y)=\displaystyle\int_0^yf_{X,Y}(x,y)\,\mathrm dx=\begin{cases}2y&\text{for }0$

Compute the expectations for both random variables:

$E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^12x(1-x)\,\mathrm dx=\frac13$

$E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\int_0^12y^2\,\mathrm dy=\frac23$

Compute the variances and thus standard deviations:

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

where

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\int_0^12x^2(1-x)\,\mathrm dx=\frac16$

$\implies V[X]=\dfrac16-\left(\dfrac13\right)^2=\dfrac1{18}\implies\sqrt{V[X]}=\dfrac1{3\sqrt2}$

$E[Y^2]=\displaystyle\int_{\infty}^\infty y^2f_Y(y)\,\mathrm dy=\int_0^12y^3\,\mathrm dy=\frac12$

$\implies V[Y]=\dfrac12-\left(\dfrac23\right)^2=\dfrac1{18}\implies\sqrt{V[Y]}=\dfrac1{3\sqrt2}$

Compute the covariance:

$\operatorname{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]$

We have

$E[XY]=\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\int_0^1\int_0^y2xy\,\mathrm dx\,\mathrm dy=\frac14$

and so

$\operatorname{Cov}[X,Y]=\dfrac14-\dfrac13\dfrac23=\dfrac1{36}$

Finally, the correlation:

$\operatorname{Corr}[X,Y]=\dfrac{\operatorname{Cov}[X,Y]}{\sqrt{V[X]}\sqrt{V[Y]}}=\dfrac{\frac1{36}}{\left(\frac1{3\sqrt2}\right)^2}=\dfrac12$